In the previous lesson, we learned about the words and, and or and intersections and unions. And was for an intersection and or was for a union. We're going to use this knowledge for compound inequalities. Let's see how these ideas play out. For a compound inequality, we don't just have one inequality, we have two. Between each compound inequality, we'll have the word and or, or. This one has and. So, for this first skills check, I want you to solve each inequality and write your answer in these boxes. Then, we're going to look at how to diagram the solution.
For the first inequality, we can just subtract 1 from both sides, and we get x is less than or equal to 8. The second inequality is a little harder. First we add 2 to both sides, then we divide by a negative 1. Remember, when I divide here, I need to reverse my inequality sign. So I get x is greater than or equal to 5. For compound inequalities though, this doesn't necessarily mean the answer. We want to figure out which numbers satisfy this condition, and this condition. Let's see if we can figure that out next.
For the first inequality, we had x was less than or equal to 8. And all of these numbers on this number line represent the solution set for that inequality. For the second inequality, we had x greater than or equal to 5. All of these numbers satisfy this condition. But we have a compound inequality, we want numbers that satisfy the first inequality and the second inequality. So, what do you think? Which number line represents solutions for both inequalities? Choose the best answer. As a hint, you can try plugging in values for different regions and see if they work in both.
So, this is the solution set for our compound inequality. If you got it right, nice work. If we stack the number lines on top of each other, like this, we can see that there's a region of intersection. These are the numbers that are shared between the inequalities. They make both of the inequalities true. And this is why we learned about intersections, and really helps us figure out what numbers make both inequalities true. We can see this region of intersection on one number line. This is the solution set. We can also write it in interval notation. So, notice that an and compound inequality makes an intersection, or a region between points. And any of these numbers would work in both inequalities. Let's check that out real quick. Let's check 6. We're going to let x equal 6 and check the solution. So, plugging in 6, I get 7 less than or equal to 9 and plugging in 6 over here, I get negative 8 is less than or equal to negative 7. Both of these statements are true. So, that means 6 is a solution to this compound inequality. Notice that for equations and inequalities, we've always been able to check our answer. This is what so great, we can know that we're right. Checking your answer can build your confidence in knowing that you're right. Or it means, hey, maybe I should go back and check my work.
Let's try this with some practice. >> What do you think the solution set is to this compound inequality? This last answer might look unfamiliar. >> We have two separate regions on the number line. >> When we want both regions, we use the simple union, because we could have these answers or these answers. >> So, what values of x make this inequality true and this inequality true? Try your best.
To solve this and compound inequality, let's solve each and inequality separately. For the first inequality, we add 4 to both sides, then divide by negative 3, to isolate x. Remember, we divided by a negative so we need to reverse the inequality sign. So, for this inequality sign, x needs to be less than negative 3. For the second inequality, we add 1 to both sides, then we divide by 5. So x is less than or equal to negative 4. We have an AND inequality, so I want both of these solutions to be true. x needs to taken a value that works for this inequality and this inequality. So I want to find the overlap of these 2 regions. Let's look back at our number lines. Here are two solution regions. x could be less than or equal to negative 4, or x could be less than negative 3. We have an and inequality, so we want the region of overlap. So, it's any number, less than or equal to negative 4. That would be this number line. Nice job if you got that one right.
Here's a different type of compound inequality. This one uses the word, or. I want any number of x that makes this inequality true, or any number of x that makes this inequality true. It doesn't have to be true for both of them. I'm going to do some leg work and go ahead and solve these inequalities. Now would be a good point to pause the video and see if you get the same solutions I do. For the first inequality, we get x is less than or equal to 1. For the second inequality, we divide by negative 3, so we reverse the inequality sign and we get x is greater than or equal to 3. Any number less than or equal to 1 or any number greater than or equal to 3 will satisfy one of these inequalities. Let's see this on a number line. So, here's our number line. Any value to the left of this inequality. Just like all the other equations and inequalities we've done, we can check our answer to make sure we're right. I'm going to check 0 for this inequality and I'm going to check 4 four for this inequality. Remember, 0 won't work for this one, but it will work for this one. We just need to make sure it will work for one or the other. If I plug in zero for x for the first inequality, I get negative 4 is less than or equal to 0. This is true. If I plug in x equals 4 into the second inequality, I get negative 12 is less than or equal to 9. This is also true. This is great. These solutions work for the first inequality, and these solutions work for the second. They don't need to work for both, they just need to work for one or the other.
Let's try using our knowledge of the word or to solve this compound inequality. Remember, you're looking for numbers for x that satisfy either the first condition or the second condition. So, what's the solution set for this compound inequality? Choose the best number line.
For the first inequality, I can add 4 to both sides to get negative 3x is greater than 9. Then we divide by negative 3, the coefficient of x, we get x is less than negative 3. Remember, we need to reverse the inequality sign here. For the second inequality, we add 1 to both sides, then we divide by 5 to get x is less than or equal to negative 4. Notice I don't need to reverse the inequality sign here because I divided by a positive. We only reverse the inequality sign when we multiply or divide by a negative. We can graph these solutions on a number line. x can be any value less than negative 3 here or x can be any value less than or equal to negative 4 here. But remember, we want x values for either this inequality or this inequality. Let's see these number lines stacked on top of each other to decide.
Here were the two original inequalities and these were the solutions, x less than negative 3 or x less than or equal to negative 4. Let's compare the two number lines. It's harder to see the number lines compared horizontally, so let's stack them on top of each other. When looking at the number lines, we values of x that satisfy this inequality or this inequality. So, we want all of these values. Any value within this region or any value within this region satisfies one of the conditions. Notice that the top number line is contained entirely within the second one. That means, we only need the solutions that are less than negative 3. Any value that's less than negative 3 will work in this first inequality and any value less than negative 4 will work in this inequality. That's why we want this region. We want these values or these values, so we show all these values on one number line. This is our solution set. Any value of x for negative infinity up to negative 3 but not including negative 3 will work. x needs to be less than negative 3. In interval notation, we list it like this. This is a really tough problem. So, if you got the one right, great work.
Here's the last compound inequality we'll look at. Notice I have two inequity signs, and an expression with a variable in the middle. So you might be wondering, is this an AND inequality, or an OR inequality. To think about it, let's try splitting it up. Here's one inequality, so I'm going to write the one here. And here's the second inequality. We want this amount to be greater than negative 2 and we want this amount to be less than 4. So this is an AND inequality. We know how to solve an AND inequality, and we could solve each of these separately, but I'm going to show you something different. We're going to solve this inequality by working with each part of the inequality. All three of them. First I'm going to clear the fraction by multiplying every part by 5. 5 times negative 2 is negative 10. Here, I'm going to put the 5 over 1, then I'm going to reduce the common factors in the numerator and the denominator. So these 5's reduce to 1. Now that you have an idea of how to perform the operations to every part, I want you to finish this one out. Write your answer in interval notation in this box. Remember to use parentheses or brackets and a comma to separate the numbers from the interval. Good luck.
When solving this inequality, we can add one to both sides to isolate the 3x. Next, to isolate x, we divide each part by 3. So, we know x is greater than negative 3 but less than 7. But this isn't our answer. You want it to be an interval notation. We know x is between these values. So, I have negative 3 and negative 3. Here's the solution on a number line. We're almost done with Unit 2. I hope so far, you've gained a stronger understanding of satisfying conditions and learning when one condition or both conditions must be met. Let's wrap up this entire section with one last lesson, a lesson with absolute value equations and inequalities.
Let's practice some compound inequalities. The first one I want you to try, is 2 x minus 1 is greater than 4 x plus 3, and 5 x plus 3 is greater than or equal to interval notation. Remember, when you're using interval notation, you want to use parentheses for greater than or less than, and square brackets for less than or equal to and greater than or equal to. Good luck.
The solution to this compound inequality is from negative 5 to negative 2. When we solve compound inequalities we need to solve each of the inequalities independently and since they're connected with an AND, we're going to need to look at the intersection of these two solutions. So lets solve this inequality first. I'm going to subtract 2x from both sides which gives me negative 1 is greater than. 2x plus 3. Subtract 3 from both sides, negative 4 is greater than important to know that negative 2 greater than x is the same as x is less than negative 2. So we have the solution to our first inequality. Let's solve our second inequality. This time I'm going to subtract 3 x from both sides. When I subtract 3 x from both sides, I get 2 x plus 3 is greater than or equal to negative 7. Subtract 3 from both sides, and I get 2x is greater than or equal to negative 10. Once again divide both sides by 2, and I get that x is greater than or equal to negative 5. Now that I've solved both inequalities, I have to see where these intervals overlap. I'm going to use a number line. I'm going to draw the solution to the two inequalities on top of this number line. X is less than negative 2 uses a parenthesis. And the solution is less than negative 2. And, for x greater than or equal to negative 5, I have a square bracket at negative 5 and my solution. Goes to the right. Since this is an AND inequality, we have to look at where these two intervals overlap. As you can see they overlap from negative 5 to negative 2, so I can draw the actual solution on the number line, which means from negative 5 to negative 2 and everything in between. Let's try another compound inequality.
Since you just got to practice a compound inequality with an and, let's try a compound inequality with a or. So, we have 6 x plus 5 is less than 3 x plus 2, or 2 x plus 7 is grater than x plus 11. We know since it's a or problem, we're going to need a union. You can put your answers in these boxes, and I've already put the union sign for you. If you need to write infinity, type in \inf and the infinity symbol should show up.
The solution to this compound inequality is negative infinity to negative 1 union 4 to infinity. How did you do? Did you get it right? If so, great work. Let's see how we got that. Once again, I'm going to solve these inequalities independently and I'm going to start skipping a step or two. When I'm solving this inequality, I want to subtract 3x from both sides, which gives me 3x plus 5 is less than 2. Subtracting 5 from both sides, gives me 3x is less than negative this inequality, I'm going to skip the steps where I'm showing what I'm adding or subtracting to each side of the equation. In other words, here, I want to subtract x from both sides. When I subtract x from 2x, I am simply left with x plus 7 on the left side and when I subtract x from the right side, x minus x is greater than 11 minus 7, which is 4. I have x is less than negative 1 or x is greater than 4. Let's see what that looks like on a number line. When I graph the solution of x is less than negative 1 on the number line, I have a parenthesis because it's less than, and I know it's going to the left because it's less than. Similarly, x is greater than 4 We've put a parenthesis at 4 and since it's greater than, we're going to the right. The solution on the number line makes it really easy to write in interval notation. Since this arrow is going off to negative infinity, we start at negative infinity and we have to stop at negative 1. Since it's an or we have a union, and then we start up again at 4 and the arrow is going to positive infinity. Let's go ahead and try one more compound inequality.
For our last problem on compound inequalities, let's try one that's written in this format. You can put your answer in this box using interval notation.
Did you get the closed interval from negative 1 to 1? If so, you should be very proud of yourself. I know I am. Let's see how we got that solution. Remember when you're working with a compound inequality that's written in this form, we can perform operations as long as we do the same operation, to all three portions of the quation. The first thing I want to do is to get rid of this fraction by multiplying through by the least common denominator, and the least common denominator is 2. When I multiply 2 by negative 3, I get negative 6. When I multiply 2 by 5 x minus 1 over 2, I'm simply left with 5 x minus 1, and when I multiply 2 times 2, I get 4. Next I'm going to add 1 to each part of the equation so I can isolate the variable in the middle. When I do that, negative 6 plus 1 is negative 5, and 4 plus 1 is positive 5. Now I can divide through by 5. Negative 5 divided by 5 is negative 1, less than or equal to x, less than or equal to 5 over 5 is positive 1. Let's draw that on the number line. This portion of the inequality says that negative 1 is less than or equal to x, which is the same as x is greater than or equal to negative 1. So, that means we're going to be going in that direction. This portion of the equation says x is less than or equal to positive 1, so we use a square bracket and the answer's going here. So we have the interval from negative 1 to 1 using the square brackets.